I'm trying to prove that Fibonacci numbers $F_i, F_{i+3}$ can be either coprime, where $i$ and $i+3$ are not both even, or they can have a greatest common divisor of $2$, when $i$ and $i+3$ are even.
I've tried using the formula that $gcd(F_m, F_n) = F_{gcd(m,n)}$, but I've not been able to come up with a full proof that ensures that the greatest common factor of Fibonacci numbers 3 terms apart have a maximum greatest common divisor of 2.
Is there any way to prove this maximum of 2, or that $F_i, F_{i+3}$ are coprime, except for when $i$ and $i+3$ are even?
Best Answer
We have $$ \begin{bmatrix} F_{n+3} \\ F_n \end{bmatrix} = \begin{bmatrix} 4 & 1\\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} F_{n} \\ F_{n-3} \end{bmatrix} $$ The matrix has determinant $-1$ and so has an integer inverse. Therefore, $\gcd(F_{n+3},F_n)=\gcd(F_{n},F_{n-3})$.
Thus it suffices to compute $\gcd(F_{3},F_{0})=2$, $\gcd(F_{4},F_{1})=1$, $\gcd(F_{5},F_{2})=1$.